The $4$-dimensional lattice $\mathbb{Z}^{4}$ has vectors of length $\sqrt{n}$ for any positive integer $n$ by the Four Squares Theorem, but this need not be true for higher-dimensional integral, unimodular lattices for two reasons:

(i) Some small positive integers could be skipped as squared lengths of lattice vectors. For example, the odd Leech lattice has no $v$ with $v \cdot v = 1$ or $2$.
(ii) The lattice may be even.

Therefore, the way to word the question to recognize these possibilities is:

Let $\Lambda$ be an integral unimodular lattice of dimension $d$, where $d \geq 4$.
(i) If $\Lambda$ is odd, then is it true that every sufficiently large positive integer arises as the squared length of a vector in $\Lambda$?
(ii) If $\Lambda$ is even, then is it true that every sufficiently large even positive integer arises as the squared length of a vector in $\Lambda$?

1 Answer
1

Yes, both are true. For example, see Theorem 1.6 of Chapter 11 of Cassels's "Rational Quadratic Forms", which says that if $Q$ is a positive definite integral quadratic form, then there is an integer $N$ depending on $Q$ such that if $a > N$ and $a$ is represented primitively by $f$ over all $\mathbb{Z}_p$ then $a$ is represented by $Q$. The local primitive representability is easy to show using the fact that $f$ is unimodular, and the classification of forms over $\mathbb{Z}_p$ by invariants.

For instance, if $p$ is odd and $p \nmid a$, then $f$ is equivalent over $\mathbb{Z}_p$ to $(a, a \det(f), 1, 1, \dots, 1)$, which obviously represents $a$ primitively. If $p | a$ you could use $((a-1), (a-1) \det(f), 1, 1, \dots, 1)$ which represents $a$ primitively. I won't do the analysis for $p = 2$, but see section 4 of chapter $8$ of Cassels.